(2 pontos ) Considere a base \left\{v_{1}, v_{2}, v_{3}\right\} de \mathbb{R}^{3}, onde v_{1}=(1,1,1), v_{2}=(1,0,1) e v_{3}=(0,0,1). Aplicando o processo de Gram-Schmidt, obtenha:
(a) uma base ortogonal \left\{w_{1}, w_{2}, w_{3}\right\};
(b) uma base ortonormal \left\{z_{1}, z_{2}, z_{3}\right\}.

Passo 1

O enunciado nos dá a base v_{1}=(1,1,1), v_{2}=(1,0,1), v_{3}=(0,0,1) de \mathbb{R}^3 e pede que a partir dela, usando o processo de Gram–Schmidt, obtenhamos
  1. uma base ortogonal \{w_{1},w_{2},w_{3}\},
  2. e então a base ortonormal \{z_{1},z_{2},z_{3}\}.
    O método consiste em, a cada passo, subtrair de v_{k} as projeções já calculadas sobre os vetores w_{1},\dots,w_{k-1}.

Passo 2

Definimos
w_{1} = v_{1} = (1,1,1).

Passo 3

Para obter w_{2}, calculamos
\mathrm{proj}_{w_{1}}(v_{2}) = \frac{\langle v_{2},w_{1}\rangle}{\langle w_{1},w_{1}\rangle}\,w_{1},
onde \langle v_{2},w_{1}\rangle = 1\cdot1 + 0\cdot1 + 1\cdot1 = 2 e \langle w_{1},w_{1}\rangle = 1^2+1^2+1^2 = 3. Logo,
\mathrm{proj}_{w_{1}}(v_{2}) = \frac{2}{3}(1,1,1) = \Bigl(\tfrac{2}{3},\tfrac{2}{3},\tfrac{2}{3}\Bigr).
Subtraindo de v_{2}, vem
w_{2} = v_{2} - \mathrm{proj}_{w_{1}}(v_{2}) = \Bigl(1 - \tfrac{2}{3},\,0 - \tfrac{2}{3},\,1 - \tfrac{2}{3}\Bigr) = \Bigl(\tfrac{1}{3}, -\tfrac{2}{3}, \tfrac{1}{3}\Bigr).

Passo 4

Para w_{3}, subtraímos de v_{3} as projeções sobre w_{1} e w_{2}:
\mathrm{proj}_{w_{1}}(v_{3}) = \frac{\langle v_{3},w_{1}\rangle}{\langle w_{1},w_{1}\rangle}\,w_{1} = \frac{1}{3}(1,1,1) = \Bigl(\tfrac{1}{3},\tfrac{1}{3},\tfrac{1}{3}\Bigr),
\mathrm{proj}_{w_{2}}(v_{3}) = \frac{\langle v_{3},w_{2}\rangle}{\langle w_{2},w_{2}\rangle}\,w_{2} = \frac{\tfrac{1}{3}}{\tfrac{2}{3}}\;w_{2} = \tfrac12\,\Bigl(\tfrac{1}{3},-\tfrac{2}{3},\tfrac{1}{3}\Bigr) = \Bigl(\tfrac{1}{6},-\tfrac{1}{3},\tfrac{1}{6}\Bigr).
Somando as projeções temos \bigl(\tfrac12,0,\tfrac12\bigr), então
w_{3} = v_{3} - \bigl(\tfrac12,0,\tfrac12\bigr) = \Bigl(-\tfrac12,0,\tfrac12\Bigr).

Passo 5

Agora normalizamos cada w_{i}. Primeiro as normas:
\|w_{1}\| = \sqrt{1^2+1^2+1^2} = \sqrt{3}, \quad \|w_{2}\| = \sqrt{\tfrac{1}{3^2} + \tfrac{(-2)^2}{3^2} + \tfrac{1}{3^2}} = \sqrt{\tfrac{6}{9}} = \sqrt{\tfrac{2}{3}}, \quad \|w_{3}\| = \sqrt{\bigl(-\tfrac12\bigr)^2 + 0^2 + \bigl(\tfrac12\bigr)^2} = \frac{1}{\sqrt{2}}.
Logo as versões unitárias são
z_{1} = \frac{w_{1}}{\|w_{1}\|} = \Bigl(\tfrac{1}{\sqrt{3}},\tfrac{1}{\sqrt{3}},\tfrac{1}{\sqrt{3}}\Bigr), \quad z_{2} = \frac{w_{2}}{\|w_{2}\|} = \Bigl(\tfrac{\sqrt{6}}{6}, -\tfrac{\sqrt{6}}{3}, \tfrac{\sqrt{6}}{6}\Bigr), \quad z_{3} = \frac{w_{3}}{\|w_{3}\|} = \Bigl(-\tfrac{\sqrt{2}}{2},0,\tfrac{\sqrt{2}}{2}\Bigr).

Resposta

(a) A base ortogonal é
\{\,w_{1},w_{2},w_{3}\} = \Bigl\{(1,1,1),\,\bigl(\tfrac{1}{3},-\tfrac{2}{3},\tfrac{1}{3}\bigr),\,\bigl(-\tfrac12,0,\tfrac12\bigr)\Bigr\}.
(b) A base ortonormal é
\{\,z_{1},z_{2},z_{3}\} = \Bigl\{\,(\,\tfrac{1}{\sqrt{3}},\tfrac{1}{\sqrt{3}},\tfrac{1}{\sqrt{3}}),\,(\tfrac{\sqrt{6}}{6},-\tfrac{\sqrt{6}}{3},\tfrac{\sqrt{6}}{6}),\,(-\tfrac{\sqrt{2}}{2},0,\tfrac{\sqrt{2}}{2})\Bigr\}.

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